# Doubts on application of continuity definition and Dominated Convergence theorem

I quote Øksendal (2003).

Let $$\mathcal{V}=\mathcal{V}(S,T)$$ be the class of functions $$f(t,\omega):[0,\infty)\times\Omega\to\mathbb{R}$$ such that $$(t,\omega)\to f(t,\omega)$$ is $$\mathcal{B}\times\mathcal{F}$$-measurable (where $$\mathcal{B}$$ denotes the Borel $$\sigma$$-algebra on $$[0,\infty)$$), $$f(t,\omega)$$ is $$\mathcal{F}_t$$-adapted and $$\mathbb{E}\bigg[\int_{S}^T f(t,\omega)^2 dt\bigg]<\infty$$.
[...] Recall that a function $$\phi\in\mathcal{V}$$ is called elementary if it has the form $$\phi(t,\omega)=\sum_j e_j(\omega)\cdot\chi_{[t_j, t_{j+1}]}(t)\tag{1}$$ [...]

Statement Let $$g\in\mathcal{V}$$ be bounded and $$g(\cdot,\omega)$$ continuous for each $$\omega$$. Then there exists elementary functions $$\phi_n\in\mathcal{V}$$ such that $$\mathbb{E}\left[\int_S^T\left(g-\phi_n\right)^2dt\right]\to 0\hspace{1.5cm}\text{as }n\to\infty\tag{2}$$
Proof Define $$\phi_n(t,\omega)=\sum_j g(t_j,\omega)\cdot\chi_{[t_j,t_{j+1})}(t)$$. Then, $$\phi_n$$ is elementary since $$g\in\mathcal{V}$$, and $$\int_S^T(g-\phi_n)^2dt\to0\hspace{1.5cm}\text{as }n\to\infty\text{ for each }\omega$$ since $$g(\cdot,\omega)$$ is continuous for each $$\omega$$. Hence $$\mathbb{E}\left[\int_S^T(g-\phi_n)^2dt\right]\to0$$ as $$n\to\infty$$ by bounded convergence.

My questions:

1. Why does definition of continuity of $$g(\cdot,\omega)$$ imply that $$\displaystyle{\int_S^T(g-\phi_n)^2dt}\to0\hspace{1.5cm}\text{as }n\to\infty\text{ for each }\omega\hspace{3.5cm}\text{?}$$

My interpretation: I think I am allowed to conceive $$\phi_n$$ as a kind of step-function, whose value at time $$t_n$$ corresponds to the value of the continous and bounded function $$g$$ at time $$t_n$$. Does that mean that if I shrink the differential of time $$[t_j,t_{j+1})$$, continuity of $$g$$ implies that $$|g-\phi_n|<\varepsilon\text{ for }|t_j-t_{j-1}|<\delta$$ (which implies that $$\displaystyle{\int_S^T(g-\phi_n)^2dt}\to0\hspace{0.5cm}\text{as }n\to\infty\text{ for each }\omega$$)?

1. In the end, is Lebesgue's dominated convergence theorem applied? If so, why does it lead from $$\displaystyle{\int_S^T(g-\phi_n)^2dt}\to0\hspace{1cm}\text{as }n\to\infty\text{ for each }\omega$$ to $$\mathbb{E}\left[\displaystyle{\int_S^T(g-\phi_n)^2dt}\right]\to0\hspace{2.3cm}\text{as }n\to\infty\text{ for each }\omega\hspace{1.8cm}\text{ ?}$$

My interpretation: What I think is that one could set $$X_n=(t_{j+1}-t_j)$$ and $$Y_n=\displaystyle{\int_S^T(g-\phi_n(t,\omega))^2}$$, which - as seen in my interpretation in point $$1.$$ - since $$g$$ is continuous, by definition of continuity, is such that for every $$t$$, $$|Y_n|<\epsilon$$ whenever $$|X_n|<\delta$$. In other words, $$g=\lim_{n\to\infty}\phi_n(t,\omega)\hspace{0.5cm}\text{ pointwise}\tag{3}$$ implies that $$0=\lim_{n\to\infty}\int_S^T(g-\phi_n(t,\omega))^2dt\hspace{0.5cm}\text{ pointwise}\tag{4}$$ Hence, given the immediately above explained conditions:

• $$|Y_n|<\epsilon\text{ for every }t$$ (namely, "boundedness"), whenever $$|X_n|<\delta$$;
• $$0=\lim\limits_{n\to\infty}\displaystyle{\int_S^T(g-\phi_n(t,\omega))^2dt}\hspace{0.5cm}\text{ pointwise}$$ (namely, "pointwise convergence")
one could apply Lebesgue's dominated convergence theorem: $$\lim_{n\to\infty}\mathbb{E}\left(Y_n\right)=\mathbb{E}\left(\lim_{n\to\infty}Y_n\right)=\mathbb{E}\left(0\right)=0$$

Are my interpretations of points $$1.$$ and $$2.$$ correct? If not, why?

1. Yes, your reasoning is correct.
For a fixed $$\omega$$, $$g(t,\omega)$$ is a continuous function of $$t$$. This means it is uniformly continuous (in $$t$$) on the compact interval $$[S,T]$$. Hence, for any given $$\epsilon > 0$$, we can find $$\delta >0$$ such that
$$|g(s, \omega) - g(t,\omega)| < \epsilon \quad \forall s, t \in [S,T]: |s-t| < \delta.$$ Now choose a $$\epsilon > 0$$ as above, and a time spacing $$S = t_0 < t_1 \ldots < t_n = T,$$ where max $$|t_i - t_{i+1}| < \delta$$.
On the every interval $$[t_i, t_{i+1})$$ we have
$$|g(t, \omega) - \phi_n(t,\omega)| = |g(t, \omega) - \phi_n(t_i,\omega)| = |g(t, \omega) - g(t_i,\omega)|< \epsilon.$$ So $$\int_S^T |g(t, \omega) - \phi_n(t,\omega)|^2 \,dt < \epsilon^2 \cdot (T-S).$$ Since epsilon was arbitrary, the integral can be made arbitrarily small by making the maximum grid spacing, max $$|t_i - t_{i+1}|$$, small enough.
You don't write what the $$n$$ in your functions $$\phi_n$$ stands for, but I assume it means the grid spacing goes to zero when $$n \to \infty$$.

2. Yes, here the Lebesgue dominated convergence theorem can be applied.
First you have to check that the integrand is uniformly bounded by an integrable function for all $$n$$. This follows from the previous point (EDIT: This is wrong, as forgottenarow below points out. You als need to use the boundedness of $$g$$ here. The subtle point here is that the n typically depends on $$\omega$$).

The integrand can be made smaller than $$\epsilon^2$$ if the grid is fine enough. This is of course an integrable function, since we consider a finite interval.
Also, for each $$\omega$$, you have pointwise convergence of $$\phi_n(t, \omega)$$ to $$g(t, \omega)$$ when $$n\to\infty$$ according to (1) (you even have uniform convergence).
So the conditions for the LDK theorem is fulfilled and you are allowed to put the limit inside the integral. So $$\lim_{n \to \infty}\int_S^T |g(t, \omega) - \phi_n(t,\omega)|^2 \,dt = \int_S^T \lim_{n \to \infty} |g(t, \omega) - \phi_n(t,\omega)|^2 \,dt = \int_S^T 0 dt = 0.$$

Regarding the expectation, you can put the limit inside by the same type of reasoning:
Let $$\mathbb{P}$$ be the probability measure in which we take the expectation. $$\begin{eqnarray*} \mathbb{E} \left[ \int_S^T |g(t, \omega) - \phi_n(t,\omega)|^2 \,dt \right] = \int \left( \int_S^T |g(t, \omega) - \phi_n(t,\omega)|^2 \,dt \right) d \mathbb{P}(\omega) \end{eqnarray*}$$ For any $$\epsilon$$ and $$n$$ large enough the inner integrand is bounded by $$\epsilon^2$$ according to 1. And, as I mentioned above, this means it is uniformly bounded for all $$n$$ by the integrable function $$\epsilon^2$$.
This function is integrable since $$\int \left( \int_S^T \epsilon^2 \,dt \right) d \mathbb{P}(\omega) = (T-S)\epsilon^2 \cdot \int d\mathbb{P(\omega)} = (T-S)\epsilon^2,$$ since the total mass of a probability measure is $$1$$.
Hence the conditions for LDK is fulfilled and you can put the limit inside the double integral and get $$0$$ in the limit as $$n \to \infty$$ as before.

• Part 1 works, but I don't think we can use the same reasoning in part 2. For every $\omega$, there exists an $n$ large enough so that the bound in part 1 holds. However, this $n$ depends on $\omega$ so it can be viewed as a random variable. This also means that for any $n$, there could be a positive probability that the integral is arbitrarily large, which would prevent the integral from converging in expectation. I think we need to use the boundedness of $g$ and $\phi_n$ here to fully justify this step. Oct 26, 2020 at 18:45
• Yes, you are right. I forgot that the n depends on the $\omega$. Oct 26, 2020 at 18:48

Note that $$g$$ is assumed to be bounded, so there exists some $$M < \infty$$ such that $$\sup_{t \in [S,T]} |g(t,\omega)| < M$$ for almost surely all $$\omega$$. Furthermore, by definition of $$\phi_n$$, $$\sup_{n \in \mathbb{N}}\sup_{t \in [S,T]} |\phi_n(t,\omega)| < M.$$

1. The almost sure continuity of $$g$$ ensures that $$|g(t,\omega) - \phi_n(t,\omega)|^2 \to 0$$ for all $$t$$ for almost surely all $$\omega$$. By boundedness, $$\sup_t |g(t,\omega) - \phi_n(t,\omega)|^2 \leq 4M^2 < \infty$$. So by bounded convergence,

$$\int_S^T|g(t,\omega) - \phi_n(t,\omega)|^2\,dt \to 0 \text{ for almost surely all } \omega\in \Omega.$$

1. Since $$\sup_t |g(t,\omega) - \phi_n(t,\omega)|^2 \leq 4M^2$$ almost surely, it follows that,

$$\int_S^T|g(t,\omega) - \phi_n(t,\omega)|^2\,dt < \int_S^T 4M^2\,dt = 4M^2(T-S)<\infty \text{ for almost surely all } \omega \in \Omega.$$

So we can apply bounded convergence again to get,

$$\lim_{n\to\infty}\mathbb{E}\left[\int_S^T|g(t,\omega) - \phi_n(t,\omega)|^2\,dt\right] = \mathbb{E}\left[\lim_{n\to\infty}\int_S^T|g(t,\omega) - \phi_n(t,\omega)|^2\,dt\right] = 0.$$

• I have some questions: could you please detail more the reasons why: 1. "almost sure continuity of g ensures that $|g(t,\omega) - \phi_n(t,\omega)|^2 \to 0$ [...]"?; 2. "By boundedness, $\sup_t |g(t,\omega) - \phi_n(t,\omega)|^2 \leq 4M^2$". Thank you a lot in advance Oct 30, 2020 at 9:26
• Sure! Let $\{t_i^n\}$ be the mesh which generates $\phi_n$. Let $\tau_n = \max\{t^n_i:t^n_i \leq t\}$. Recall that $\phi_n(t) = g(\tau_n)$. However, since the mesh size (maximum value of $|t^n_{i+1}-t^n_i|$) goes to zero with $n$, it follows that $\tau_n \to t$. Thus, $\lim_{n\to\infty} \phi_n(t,\omega) = \lim_{n\to\infty} g(\tau_n,\omega) = g(t,\omega)$ by a.s. continuity of $g$. For your second question, since $g$ and $\phi_n$ are both bounded by $M$, $$|g - \phi_n|^2 \leq \left(|g| + |\phi_n|\right)^2 \leq (M+M)^2 = 4M^2.$$ Oct 30, 2020 at 18:38